BinSearch | Super-efficient , in-memory key/value data store | Key Value Database library
kandi X-RAY | BinSearch Summary
kandi X-RAY | BinSearch Summary
binsearch is a super-efficient, in-memory key/value data structure for go. in future it could also expand to be disk-based easily enough. the different structure names are one of key, keyval, counter, followed by one of bytes, runes, int, uint64, uint32, uint16, uint8. e.g. keyvalbytes, counteruint32. key and keyval types should never have duplicate keys added. it is important to either use find(key) to check if a key exists before adding it (see example 4), or to use the counter structure to remove duplicates first (see example 10). most structures require the keys to be added first and then the build() function executed before any find(key) is performed. the exception to this is the add(key) function from key and keyval types, which does not require the use of build() and which does allow for find(key) to be performed at any time, but the insertion of the keys is considerably slower. in most cases this is not necessary and there is usually a way to avoid using add(key). the key store records only the keys and its find(key) function will return an index number, which is the position of the key. you can store any value in a separate slice under this
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Top functions reviewed by kandi - BETA
- uint62bytesend appends a uint64 to word .
- doPivotAsc performs pivot operation .
- runes2bytes converts a slice of runes into a byte slice .
- bytes2uint64 converts a byte slice to an uint64
- bytes2runes converts a byte array to a byte slice .
- quickSortAsc performs an ascending sort .
- ssiftDownAsc reduces the data in ascending order .
- Build returns a sorted sorted list of keys .
- swapSortAsc sorts the data in ascending order .
- medianOfThreeAsc orders the median of two arrays .
BinSearch Key Features
BinSearch Examples and Code Snippets
Community Discussions
Trending Discussions on BinSearch
QUESTION
Array extend [
Array class >> bin: val left: l right: r [
^ super binSearch: val left: l right: r
]
binSearch: val left: l right: r [
|arr iter|
arr := self.
[l == r]
ifTrue: [^ (-1)].
iter := (r + l)/2.
[arr at: iter == val]
ifTrue: [^iter].
[arr at: iter < val]
ifTrue:[^ super binSearch val left: l right: iter]
ifFalse: [^ super binSearch val left: iter right: r]
]
]
arr := #(3 6 9 10).
arr bin: 6 left: 1 right: 4.
ANSWER
Answered 2021-Jun-05 at 09:51The first part makes no sense to me. Your algorithm is recursive and so it should call itself rather than delegating to the class, which is a different object.
To do this, replace super with self and the same method will be invoked again with the new parameters.
QUESTION
I'm trying to return the index of the element to be searched. All the elements are being detected except the first element. What's wrong? Why is the function not returning index 0?
...ANSWER
Answered 2021-Jan-17 at 07:14int low=a[0];
int high=a[x-1];
QUESTION
def binsearch(a):
if len(a) == 1:
return a[0]
else:
mid = len(a)//2
min1 = binsearch(a[0:mid])
min2 = binsearch(a[mid:len(a)])
if min1 < min2:
return min1
else:
return min2
ANSWER
Answered 2020-Oct-27 at 03:57If min1 and min2 are numbers, and there's always 2 (a constant) of them, the amount of work on that particular line (a single comparison) never changes. Therefore its time complexity is O(1).
What may change, however, is the number of times that line is executed! When you have n times an O(1) operation, the overall time complexity is still O(n).
QUESTION
Supposed you've got a large amount of boxes drawn, and the user can draw a rectangular area over them.
While I'll be implementing it inside a browser, let's abstract it away and say we've got the coordinates of every point of every rectangle.
What are the most efficient data structures and algorithms here, given I want to check which boxes a) intersect b) are contained by the selection?
My current idea is to:
- Sort all boxes by
x - Via binsearch, check which boxes overlap x-wise with the selection area, then, for every x-wise overlapping box, check if they align y-wise as well.
or
- Sort all boxes by
xandy, each in separate array - Via binsearch, first find all x-overlapping boxes, then all y-overlapping boxes, then check which boxes are in both sets,
... though I'm pretty sure there's some well-known algorithm for such a problem.
...ANSWER
Answered 2020-Aug-28 at 13:40I suppose by selected via some rectangle you mean either intersects some rectangle or is contained in some rectangle. If the "drawn boxes" are of fixed position, one approach which comes to mind is binary space partition. Roughly speaking, an (ideally balanced) binary space partition tree could be generated for the "drawn boxes". If the selection rectangle is positioned, the positions of its corners would be matched against the binary space partition tree, and large halfspaces could be excluded from explicit checking for intersection.
QUESTION
PROBLEM: Given integers n, k (1 <= n <= 100.000), (1 <= k <= n). Find the length of the longest substring with at most K repeating characters and find position of the first element in the substring from the original string.
EXAMPLES
input: 6, 2 'abbbaa'
output: 4 3 //here, 4 is max length and 3 is the position of the first element of substring 'bbaa' in the original string. Thus, each character repeats k times here, so it`s OK
input: 3, 1 'abb' output: 2, 1 //yeah, the required substring is 'ab' I hope the statement is clear, at least I tried, `cause English is not my native language.
Don`t know how to proceed in designing the algorithm for this problem. I should use binary search the answer technique, as it seems from the statement of the task.
First things first: I tried at least to develop my thoughts on my way of tackling this problem, so it is:
- Find the middle, and we cut the string in two slices from 0 to mid+1 and mid to n respectively. (I code in Python, so it`s very natural to use slices here, I think)
- Look for the number of unique characters (I used len(set(slice))) and compare: if the number of 1-st (left) slice is bigger then we change the right = mid, else left
- All this instructions are put in the 'while right > left + 1' loop.
So far so good, but probably I missed sth or this idea isn`t an efficient one to get the position of the first element of the substring looking at its position in given string.
Code (Python 3): to find the position of the first element, using binsearch
...ANSWER
Answered 2020-Jul-19 at 09:29Your approach is greedy and won’t get right results. Consider input 4 4 ccab - your binary search code will return 2, but it should return 1 (whole string is good). To solve this problem you can use method called two pointers. We define two pointers a and b as the beginning and end of substring that we are currently considering. We also need to remember how many occurences of each letter there are in current substring. First, set a and b to 0. Then try to expand current substring by moving its end by 1 (b=b+1) and as we do that increment the number of occurrences of letter s[b] by 1. If this number goes beyond K this means that our current string is no longer valid. To make it valid again we need to move it's beginning to the point where all letters of substring (a, b) occur less than or exactly K times. To do this we simply increment a by 1 and decrement oc[s[a]] until oc[s[b]] <= k. Time complexity is linear, because we increment b exactly n times and a at most n times. At each step we also check if b-a+1>max_len if so set max_len to b-a+1 and the beginning to a.
QUESTION
First things first, let me tell you how I am actually using the algorithms, lest any confusion arises, since this is NOT the typical use of a Binary Search algorithm.
I needed the functions for purely academic / testing purposes to compare speeds against other functions I've tested, including Python's in-built isinstance() and in methods / operands.
I'm going through a string that contains digits and letters and testing whether a character from it is an integer.
Thus, the application of the following algorithms upon each iteration (i.e. character) of the loop upon the string is thus:
binSearch("0123456789", "4", 0, 10)
The "4" is just as an example as you can tell.
binSearch2("0123456789", "w")
The "w" is just as an example as you can tell.
NOTE: I am aware of the existence of the bisect module. That's not the point and object of my experiment and exercise, though.
#RECURSIVE VERSION BELOW
...ANSWER
Answered 2020-Jul-10 at 18:16Both of your functions can be improved by making sure that mid is excluded from the range of indexes that you consider on the next pass (either a recursive call, or a subsequent loop). Since you just checked if the mid index has the value you want (and it didn't), you can exclude mid from the upper interval, by setting low to mid+1 on the next pass. It's already being excluded from the lower intervals, since the up index is beyond the end of the interval (it's half-open).
You've also hard-coded a strange base failure case into the function, where you check for mid==9. That won't work properly for a whole lot of inputs (such as shorter strings), and may cause your code to run forever, or raise exceptions depending on where the needle character is relative to the characters in the haystack string (try searching for ' ' with your current code). The proper base case test is low >= up, which indicates that your search interval is empty. For the iterative code, you'd keep looping as long as low < up.
Here's how I'd update your functions:
QUESTION
I have tried implementing the divide and conquer technique of binary search using recursion. The code for it can be seen below. I think when the program is run, I'm getting stack overflow. If anyone does manage to find a solution, I would greatly appreciate a reason for why this happens.
...ANSWER
Answered 2020-Jun-05 at 20:31You're calling binSearch but never returning the value of that call. I think you're trying to treat found as a static variable in C, but there's no such thing in Java. Even if you modify it in another stackframe, the value in your current stackframe will still stay false.
A better way would be to use tail-recursion and directly return the result.
QUESTION
I am learning C at the moment and was given this code in a book for doing a binary search. I am still very confused on calling a C function with arguments and the book so far hasn't given me much context. I believe the 3rd argument is memory storage (i'm not entirely sure). I tried google/bing but many examples have a main function along with another function which is being called in by main. What am I missing? I tried to call binsearch like I do in Python but i've gotten a bunch of errors.
...ANSWER
Answered 2020-May-27 at 00:24Basically, you need to do something like this. But as comments recommends, (even though it is easy to skip ahead when we already know some other language) please go back and make sure you pick up the basics of C first.
QUESTION
Background:
I have a service where I have an open layers based map. The map component has its own service.
A pop up is generated when an icon is clicked. On this pop up is a button which when clicked takes some data from the object displayed on the pop up and then navigates to a different component.
Im trying to publish/add this data to a behavior subject and then subscribe to it in the new component. Data is added to the behavior subject when the icon is clicked. The pop up component has its own service.
In the destination component I'm trying to subscribe to the pop up services behavior subject in the template using an async pipe.
Problem:
The observable on the behavior subject only returns the initial value. The next function on the behavior subject does not seem to be adding new data to it. It always gives the new value.
When Replay subject is used, nothing shows. Next function simply appears to have no effect at all.
Code :
Popup Service
...ANSWER
Answered 2020-Apr-07 at 06:13I think what happens is that you create a new observable everytime getElement$() is being called with the asObservable() function. I feel the docs are not 100% clear on this.
What I suggest is to create the observable on class level as well instead of on method level.
So in your Popup service you can do this:
QUESTION
I have made two Python functions below, one for sequential (linear) search and other for binary search.
I want to do these 3 things for each size value in the given list :
generate a list of random integer values (between 1 to 1000,0000) for a given list size
run a sequential search for -1 on the list and record the time elapsed by sequential search
run a binary search for -1 on the sorted list (after sorting the list), and record the time elapsed by binary search
What I have done is :
...ANSWER
Answered 2018-Nov-06 at 07:161) You need to account for the sorting time. Binary search works only on sorted lists so sorting takes time, which takes the time complexity for it to O(nlogn). And in your case you are sorting it after the timer has started, So it will be higher.
2) You are searching for an element that doesn't exist in the list i.e -1 which is not the average case for Binary Search. Binary search's worst case has to make so many jumps just to never find the element.
3) Please do not use list as a variable it is a python's keyword and you are clearly overriding it. Use something else.
Now if you sort the list without timing it. Results change drastically. Here are mine.
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